How to calculate the wind force acting on a water drop?

[Lotto]

TL;DR Summary: I have a water drop falling with a constant velocity $v$, since $mg=\frac 12 C \rho S v^2$. Wind is blowing with a velocity $u$ only in a horizontal direction. What will be its force acting on the drop?

I would use this equation $F= \frac 12 C\rho S u^2$. When I want to calculate the net velocity of the drop, will it be the sum of $\vec v$ and $\vec u$? Bacause we assume that the drop is so light that its horizontal velocity will be equal to the wind's one.

So the drop's velocity should be then constant, but when threre is a force $F$ acting on it, how is it possible? Doesn't act on the drop also a force $-F$, the force caused by a normal air resistance?

 

[vanhees71]

I would guess an ansatz for the drag force like
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}|(\vec{v}-\vec{u}),$$
where $\vec{u}$ is the velocity field of the air. Of course after a sufficiently long time the drop gets into equilibrium with the flow of the air, and then
$$\vec{F}_{\text{drag}}-m \vec{g}=0,$$
if $\vec{u}=\text{const}$.

 

[erobz]

In the vertical direction the drop is always imparting vertical momentum to the air as it falls. In the horizontal direction the air is imparting momentum to the drop, and in the limit as the drops horizontal velocity approaches $u$ ( the speed of the air mass) the horizontal force goes to zero.

 

[Lotto]

In the vertical direction the drop is always imparting vertical momentum to the air as it falls. In the horizontal direction the air is imparting momentum to the drop, and in the limit as the drops horizontal velocity approaches $u$ ( the speed of the air mass) the horizontal force goes to zero.

And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?

 

[erobz]

And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?

Imagine you are running as fast as you can with a friend in front of you also running that fast at an arms length ahead of you ( maybe you pushed them to that speed, maybe you didn’t…it doesn’t matter) You reach out your arm, can you accelerate your friend? Also there is another friend in front of them jogging backwards at an arms ahead of your friend, can they accelerate your friend slowing them down while also moving at that speed?

 

[Baluncore]

And why does it go to zero when it approaches that velocity? Is that because of a drag force in a horizontal direction that is equal to the wind force?

It is because the horizontal velocity of the droplet is then equal to the wind velocity, so there is no differential velocity to square in the drag equation.

 

[A.T.]

So the drop's velocity should be then constant, but when threre is a force $F$ acting on it, how is it possible? Doesn't act on the drop also a force $-F$, the force caused by a normal air resistance?

There is only one force by the air on the drop, and its direction is parallel to the flow of the air relative to the drop (ignoring lift).

 

[Chestermiller]

The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where $V=\sqrt{u^2+v^2}$. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$

 

[Lotto]

The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where $V=\sqrt{u^2+v^2}$. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$

So when the wind accelerates the drop in the horizontal direction, can I write for the "final" time $t$:$\frac 12 C \rho Su^2 t=mu$? That time corresponds to the time when the drop gains the velocity $u$.

 

[Chestermiller]

Not even close. The correct differential equation, assuming no gravity, is $$m\frac{du_{drop}}{dt}=\frac{1}{2}C\rho S(u_{air}-u_{drop})^2$$Do you know how to solve this ordinary differential equation?

 

[Lotto]

Not even close. The correct differential equation, assuming no gravity, is $$m\frac{du_{drop}}{dt}=\frac{1}{2}C\rho S(u_{air}-u_{drop})^2$$Do you know how to solve this ordinary differential equation?

I think I cannot solve it, so what is the solution? Because in this problem below,

I used exactly that my equation above and I got a correct solution. So doesn't the solution correspond to that equation?

 

[A.T.]

So when the wind accelerates the drop in the horizontal direction, can I write for the "final" time

For the problem above you don't need time. Just assume terminal velocity, where drag balances gravity, and horizontal velocity equals wind-speed. In the rest frame of the airmass the drop just falls vertically, so the drag is only vertical.

 

[Lotto]

For the problem above you don't need time. Just assume terminal velocity, where drag balances gravity, and horizontal velocity equals wind-speed. In the rest frame of the airmass the drop just falls vertically, so the drag is only vertical.

Yes, I solved it this way as well, but then I wanted to solve it using that wind force and I used that equation in #9 and I got a correct solution. So why could I use it when it is wrong?

 

[Chestermiller]

I would be interested in seeing the derivation of the "accepted solution" to this problem (because I'm not sure it would be consistent with the correct equations I wrote). I would like to evaluate the approximations that were made to arrive at the accepted solution.

 

[Lotto]

I would be interested in seeing the derivation of the "accepted solution" to this problem (because I'm not sure it would be consistent with the correct equations I wrote). I would like to evaluate the approximations that were made to arrive at the accepted solution.

My solution:

The "official" solution:

I have the same result, so how is that possible when my main equation is wrong?

 

[pbuk]

My solution:
View attachment 328443
The "official" solution:
View attachment 328445
I have the same result, so how is that possible when my main equation is wrong?

It is hard to say because your handwriting is difficult to read, however it seems likely that you have made a compensating error. Given that the correct equation is $\frac 12 C \rho Su^2 = mg$ whereas you have $\frac 12 C \rho Su^2 = m \frac ut$ I suspect that you have somehow managed to use a numerical value for $t$ that makes $\frac ut = g$.

 

[Chestermiller]

My solution:
View attachment 328443
The "official" solution:
View attachment 328445
I have the same result, so how is that possible when my main equation is wrong?

What this solution indicates and the problem statement fails to mention is that the droplets are traveling with the air velocity in the horizontal direction, such that $$u=u_{air,x}-u_{drop,x}=0$$Also, the velocity of the air in the vertical direction is zero, such that $$v=0-v_{drop,y}=-v_{drop,y}$$, and $$V=\sqrt{u^2+v^2}=v_{drop,y}^2$$So $$F_x=0$$$$u_{drop,x}=u_{air,x}$$and $$F_y=\frac{1}{2}C\rho Sv_{drop,y}^2=mg$$

 

[vanhees71]

The drag force is based on the relative velocity of the air with respect to the drop (not the absolute air velocity), and acts in the direction of the air with respect to the drop, where $V=\sqrt{u^2+v^2}$. So $$F=\frac{1}{2}C\rho SV^2$$The horizontal and vertical components of the drag force are $$F_x=\frac{1}{2}C\rho SuV$$ and $$F_y=\frac{1}{2}C\rho SvV$$

Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where $\vec{v}$ is the body under consideration and $\vec{u}$ is the fluid flow field.

 

[pbuk]

Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where $\vec{v}$ is the body under consideration and $\vec{u}$ is the fluid flow field.

We have reached equilbrium so $\vec u = \vec 0$.

What this solution indicates and the problem statement fails to mention is that the droplets are traveling with the air velocity in the horizontal direction.

To be fair the OP did mention this:

[The drop's] horizontal velocity will be equal to the wind's one.

 

[Chestermiller]

Forces and velocities are vectors! The correct law for a drag force quadratic in the velocity should read
$$\vec{F}_{\text{drag}}=-\alpha |\vec{v}-\vec{u}| (\vec{v}-\vec{u}),$$
where $\vec{v}$ is the body under consideration and $\vec{u}$ is the fluid flow field.

I did treat them as vectors. But it is important to recognize that the drag force vector is proportional to the relative velocity vector between the air and the drop. In the OP, the member treats u and v as the absolute air velocities in the heretical and horizontal directions.

 

[Lotto]

I did treat them as vectors. But it is important to recognize that the drag force vector is proportional to the relative velocity vector between the air and the drop. In the OP, the member treats u and v as the absolute air velocities in the heretical and horizontal directions.

OK, I understand that this equation for the horizontal velocity $\frac 12 C \rho S u^2t=mu$ is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?

I understand the "official' solution but I wanted to solve it differently. But the steps should be similar, shouldn't they? I need to express the velocity $u$ using time $t$ as I did it in my solution, then calculate the time and by using it express the unknown velocity.

However, I still don't understand how it is possible that I got a correct solution when my equation for the horizont velocity was wrong. Is it only a coincidence? But it is a weird coincidence.

 

[A.T.]

I need to express the velocity $u$ using time $t$...

Why do you need to use time here at all? Are you trying to solve a different problem, where the drops are released at rest and then are accelerated by the wind?

 

[Lotto]

Why do you need to use time here at all? Are you trying to solve a different problem, where the drops are released at rest and then are accelerated by the wind?

As it has been mentioned somewhere above, the wind accelerates the drops until they have the same velocity like the wind, so I wanted to solve it with this assumption. The "official" solution has an asumption that the water drops have been already accelerated to the velocity $u$.

 

[pbuk]

OK, I understand that this equation for the horizontal velocity $\frac 12 C \rho S u^2t=mu$ is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?

You can't: there is no horizontal drag force because there is no horizontal relative velocity.

However, I still don't understand how it is possible that I got a correct solution when my equation for the horizont velocity was wrong. Is it only a coincidence? But it is a weird coincidence.

Not a coincidence, a compensating error. The key to the solution is understanding that the ratio of the horizontal and vertical components of velocity is determined by the fact that the raindrops hit Robo's shoulder, and that the vertical component of velocity is simply terminal velocity. The equation you wrote down incorrectly for horizontal drag force (which is of course 0) was nearly the right one for (vertical) terminal velocity, and the way you used it to calculate time compensated for this.

 

[pbuk]

As it has been mentioned somewhere above, the wind accelerates the drops until they have the same velocity like the wind, so I wanted to solve it with this assumption. The "official" solution has an asumption that the water drops have been already accelerated to the velocity $u$.

But where does time come into this? Do you understand that the drops do not reacu $u$ in finite time?

Again you have got the right numerical answer because you have calculated $u = \frac xt$ and $v = \frac yt$ so your meaningless value of $t$ has cancelled out of the answer given by $\frac u v$

 

[Lotto]

You can't: there is no horizontal drag force because there is no horizontal relative velocity.Not a coincidence, a compensating error. The key to the solution is understanding that the ratio of the horizontal and vertical components of velocity is determined by the fact that the raindrops hit your shoulder, and that the vertical component of velocity is simply terminal velocity. The equation you wrote down incorrectly for horizontal drag force (which is of course 0) was nearly the right one for (vertical) terminal velocity, and the way you used it to calculate time compensated for this.

When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using $\frac 12 C \rho Sv^2=mg$. That I understand. But when the wind will suddenly start to blow with the velocity $u$, it has to accelerate the drops in a horizontal direction to the velocity $u$, because until the wind and the drops have the same velocities, the wind force is not zero. That's how I understand it.

So my attempts are to find an equation connecting $u$ and $t$.

 

[pbuk]

When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using $\frac 12 C \rho Sv^2=mg$. That I understand. But when the wind will suddenly start to blow with the velocity $u$, it has to accelerate the drops in a horizontal direction to the velocity $u$, because until the wind and the drops have the same velocities, the wind force is not equal. That's how I understand it.

The wind does not suddenly start to blow, the wind has been blowing the whole time the drop has been falling for 1000s of metres. In this time, the drops have reached a horizontal velocity of ($u - \epsilon)$ where $\epsilon$ is negligible. We are not concerned with this, we are only concerned with what happens between the roof and Robo's shoulder.

 

[Lotto]

The wind does not suddenly start to blow, the wind has been blowing the whole time the drop has been falling for 1000s of metres. In this time, the drops have reached a horizontal velocity of ($u - \epsilon)$ where $\epsilon$ is negligible. We are not concerned with this, we are only concerned with what happens between the roof and your shoulder.

I know that the wind has been already blowing, I just wanted to show what I want to do - to calculate the time $t$ needed to accelerate the drops to the velocity $u$, because the wind had to accelerate them, no matter what time exactly.

 

[erobz]

When there was no wind, the drop would be falling only in a vertical direction and then its velocity would be determined by using $\frac 12 C \rho Sv^2=mg$.

Thats its terminal velocity. Its only approached in the limit as $t \to \infty$

In the absence of "wind" ( 1D motion) the velocity as a function of time $v(t)$ is expressed as the solution to the nonlinear ODE:

$$ \uparrow^+ \sum F = \frac{1}{2} \rho A C_d v^2 - mg = m \dot v $$

 

[pbuk]

I just wanted to show what I want to do - to calculate the time $t$ needed to accelerate the drops to the velocity $u$

Do you understand that the drops do not reach $u$ in finite time?

 

[Lotto]

And why don't they reach $u$ in finite time?

 

[pbuk]

And why don't they reach $u$ in finite time?

To give you a helpful answer to this we need to know a bit more about your background in Physics and what level you are studying at.

Edit: oh and I've just realised this is not in the Homework section where it belongs, I'll get it moved.

 

[Lotto]

To give you a helpful answer to this we need to know a bit more about your background in Physics and what level you are studying at.

Edit: oh and I've just realised this is not in the Homework section where it belongs, I'll get it moved.

At the beginnig, it wasn't supposed to be a homework, my question was inspired by it.

And I studied at a high school, now I am going to go to university.

 

[erobz]

And why don't they reach $u$ in finite time?

Because that is a characteristic of the solution to the proposed ODE.

$$ \uparrow^+ \sum F = \frac{1}{2} \rho A C_d v^2 - mg = m \dot v $$

$$ \implies \dot v = \frac{1}{2 m} \rho A C_d v^2 - g $$

as $v$ increases $\dot v \to 0$. That creates an issue that is circular, because in order for $v$ to increase we need $\dot v > 0$. The acceleration $\dot v$ keeps getting smaller, and the change in velocity keeps getting smaller add infinitum as $t \to \infty$

 

[Chestermiller]

OK, I understand that this equation for the horizontal velocity $\frac 12 C \rho S u^2t=mu$ is wrong because the velocity in the drag formula is relative and when the drop has a wind's velocity the force is zero, but how to calculate the wind's velocity using the drag force formula for the horizontal direction?

They are telling you that the drop velocity is equal to the air velocity by the time that the air flows under the shelter. That means that the horizontal component of the drag force has dropped to zero by the entrance to the shelter and the drop velocity equals the air velocity. For this problem, the time that a parcel of air with its entrained water drop enters the shelter is taken as t = 0.

Therefore, you have $$u_{air,x}t=R$$and $$v_{drop,y}t=H-h$$where H is the height of the shelter overhang and h is the height of the top of the guy's head. Eliminating t between these equations gives $$u_{air,x}=\frac{R}{H-h}v_{drop,y}$$

 

[A.T.]

... the wind accelerates the drops...

Only if the wind suddenly changes. If you have a uniformly moving airmass, then the drops are created within it already moving at the same horizontal velocity as the airmass.

The horizontal acceleration of the drops and the time it takes are completely irrelevant to this problem with uniform wind.

 

[jbriggs444]

What follows will be an attempt to show how it takes infinite time for the drop to reach the horizontal velocity of the wind. This will be done without using differential calculus. I have done this derivation in the past, but goodness knows where it is buried.

We will work in one dimension. The two dimensional situation is rather nasty because vertical relative velocity affects horizontal drag force and horizontal relative velocity affects vertical drag force. Let us take a step back and consider this "two dimensional complication" in greater detail.

1. Constant drag.

This is the situation for standard kinetic friction. Force magnitude (for a given normal force) is a constant.

Possibly you have seen cars out in the parking lot in snowy weather "doing doughnuts". With rear wheel drive automobiles, this is done by stepping on the accelerator so that the back wheels spin rapidly while the car accelerates slowly forward. The driver steers hard in one direction. The front wheels stick while the rear of the car spins around the front. The key here is that the interaction between ice and tires is "constant drag".

The magnitude of the frictional force between rear tire and ice is fixed. At a high relative speed (rapidly spinning tires), the direction of this force is almost purely forward. The lateral component is negligible. So the back end can slip sideways while the front end does not.

With constant drag, a high slip rate in one direction means low drag in the perpendicular direction.

2. Linear drag.

Here, force is directly proportional to relative velocity.

Linear drag occurs when drag is primarily the result of viscosity rather than turbulence and wave effects. This will tend to be the case for very viscous fluids, low speeds and small objects. One can summarize these factors into a number called the "Reynolds Number". If the Reynolds number is low, drag is approximately linear. If the Reynolds number is high, drag tends to be quadratic instead.

With linear drag, horizontal force depends only on horizontal relative velocity and vertical force depends only on vertical relative velocity.

You can solve for the horizontal and vertical components independently. This is convenient.

3. Quadratic drag.

Quadratic drag is the usual situation for human scale interactions. Force is proportional to the square of relative velocity. So we have a force law such as $F=kv^2$ where the $k$ includes the coefficient of drag, the fluid density and any unit conversion factors.

Let us work an example. Set $k=1$ for convenience. We have a raindrop moving with a horizontal velocity of 3 and a vertical velocity of 0. What is the resulting force? $F=kv^2 = 9$. Here, $F$ is in a purely horizontal direction so $F_x = 9$.

Now take the same raindrop but give it a vertical velocity of 4. What is the resulting force now? $F=kv^2 = 25$. This time the force is split into components with $\frac{3}{5}$ in the horizontal direction and $\frac{4}{5}$ in the vertical direction. The horizontal component is $F_x = 25\frac{3}{5} = 15$.

With quadratic drag, a high slip rate in one direction means high drag in the perpendicular direction.

Back to the derivation. Clearly it will be much easier if we stick to one dimension.

Let us use the frame of reference where the air is at rest and the projectile/raindrop is moving. We want to ask "how much time will elapse before the raindrop comes to rest".

Let us say that the raindrop starts with velocity $v$. How much time will elapse before the raindrop is slowed to $\frac{v}{2}$?

Well, the drag force on the raindrop during this time interval will be at least as much as the drag force at relative velocity $\frac{v}{2}$. So we divide that force by the raindrop's mass and get an answer. It does not matter much what that answer is. Call it $\Delta t$.

The time required to cut the raindrop's velocity in half is at least $\Delta t$.

How much time is required to cut the raindrops velocity in half again?

This time we only have to lose half as much speed. If we look at the drag force at the end of the interval, it will be the result of a relative velocity of $\frac{v}{4}$. That's half the velocity to lose and one quarter of the force. It will take twice as long.

The time required to cut the raindrop's velocity in half again is at least $2 \Delta t$.

We can keep going. The total time required to reduce the raindrop's velocity to zero is at least the sum:$$T = \Delta t \times \sum_{i=0 \to \infty}2^i = \infty$$That's infinite. It takes infinite time for the raindrop to come to rest under quadratic drag.

We could also ask about how much distance is covered during the decelleration.

It will turn out that the distance covered for each halving of velocity will be a constant. So we get that the distance is at least:$$S = \Delta s \times \sum_{i=0 \to \infty} 1 = \infty$$So that is also infinite.

If one shifts to linear drag, the situation changes somewhat. It still takes infinite time -- it's the infinite sum of a constant series. That diverges to infinity. But it takes finite distance -- it's the infinite sum of a decaying geometric series. That converges to a finite sum.

With constant drag, the SUVAT equations immediately tell us that both elapsed time and distance travelled are finite.

We are not done yet...

Let us step back to the two dimensional case. This time we will be considering a raindrop that is (at least approximately) at terminal velocity in the vertical direction. It has some non-zero relative velocity $v$ in the horizontal direction.

Clearly, the magnitude of the resultang velocity will quickly (geometrically) converge on terminal velocity with a drag force of approximately $mg$. The horizontal component of drag will depend almost entirely on the angle of descent.

With the small angle approximation, $F_x = mg \sin \theta = mg \frac{v_x}{V}$ where $\theta$ is the angle of descent (measured from the vertical) and $V$ is terminal velocity.

This is effectively linear drag in the horizontal direction.

We've already reasoned above that it takes infinite time and finite horizontal distance to bring the raindrop to horizontal rest relative to the wind.

 

[haruspex]

You seem to have used “slip rate in one direction" differently in these two conclusions.

With constant drag, a high slip rate in one direction means low drag in the perpendicular direction.

I think you mean that if the slip (all of it) is in one direction then there is no drag in the perpendicular direction. Merely having a high slip rate in a given direction does not mean there is no slip (perhaps greater) in another direction.

With quadratic drag, a high slip rate in one direction means high drag in the perpendicular direction.

Here you are assuming there is slip in the perpendicular direction. How about “With quadratic drag, if the slip has components in each of two perpendicular directions then increasing the slip rate in one increases the drag in the other"?

 

[jbriggs444]

Here you are assuming there is slip in the perpendicular direction.

Yes, indeed. The alternative seems to be one dimensional motion which is the easy case.

 

[vanhees71]

The only case you can solve analytically is the "free fall". The equation of motion corresponding component of $v$ (with the gravitational acceleration in the negative direction of the corresponding basis vector) reads (assuming that for the initial condition $v(0)=0$ we always have $v<0$)
$$m \dot{v}=-m\alpha |v| v-g=m \alpha v^2 -m g$$
or
$$\dot{v}=\alpha v^2 -g.$$
This equation can be solved by "separation of variables":
$$\frac{\mathrm{d} v}{\alpha v^2-g}=-\mathrm{d} t.$$
Integrating gives
$$t=-\frac{1}{\sqrt{\alpha g}} \mathrm{artanh} \left (\sqrt{\frac{\alpha}{g}} t \right ),$$
where I've assumed that at $t=0$ we start with $v=0$. Solved for $v$ gives
$$v=-\sqrt{\frac{g}{\alpha}} \tanh \left (\sqrt{\alpha g} t \right).$$
For $t \rightarrow \infty$ we get $v\rightarrow -\sqrt{g/\alpha}$. So the motion goes asymptotically into uniform motion, i.e., the velocity adjust such that the friction force and the gravitational force compensate each other.

Integrating wrt. $t$ assuming the initial condition $x(0)=0$ you get
$$x=-\frac{1}{\alpha} \ln \left [ \cosh(\sqrt{g \alpha} t) \right]$$
For $t \rightarrow \infty$ one has $\cosh(\sqrt{g \alpha} t) \rightarrow \exp(\sqrt{g \alpha} t)/2$ and thus
$$x \simeq -\sqrt{\frac{g}{\alpha}} t + \frac{1}{2 \alpha} \quad \text{for} \quad t \rightarrow \infty.$$